I'm trying to follow Day's argument to prove that $[\mathbf C,\mathbf{Sets}]$, where $\bf C$ is symmetric monoidal, is itself symmetric monoidal, but I'm stuck at the very beginning. Is there a way to prove that the convolution of two functors defines an associative "tensor" on $Psh(\mathbf C)$?

I apologize if the question seems too boring, please notice that

  1. Day's original paper is out of reach where I am;
  2. Google seems to give me nothing useful,
  3. As you can see, the nlab page lacks the proof of associativity.

If I try to write down $F\star G\star H$ in the two ways, I feel I have to exploit some kind of Yoneda lemma written in coend form, but I'm not such an expert on ''endy'' notation to find such a form out.


  • 1
    $\begingroup$ BTW, the natural monoidal structure is induced on $[\mathcal{C}^{op}, \mathbf{Set}]$ not on $[\mathcal{C}, \mathbf{Set}]$. $\endgroup$ – Michal R. Przybylek Jan 31 '13 at 17:56

You are right that the Yoneda lemma plays the key role in the proof.

By the definition of convolution: $$(F \otimes G) \otimes H \approx \int^{C, D} \left(\int^{A, B} F(A) \times G(B) \times \hom(C, A \otimes B)\right) \times H(D) \times \hom(-, C \otimes D)$$ Because products in $\mathbf{Set}$ preserve coends, the above is isomorphic to: $$\int^{A, B, C, D} F(A) \times G(B) \times \hom(C, A \otimes B) \times H(D) \times \hom(-, C \otimes D)$$

The Yoneda lemma says that any covariant functor $K(A)$ is naturally isomorphic to $\hom(\hom(A, -), K)$. By definition, the last object is the end $\int_C K(C)^{\hom(A, C)}$. Therefore the Yoneda lemma says: $$K(A) \approx \int_C K(C)^{\hom(A, C)}$$ By duality, we get the Yoneda lemma for contravariant functors $K$: $$K(A) \approx \int^C K(C) \times \hom(A, C)$$ and from the perspective of the opposite category, for covariant functors $K$: $$K(A) \approx \int^C K(C) \times \hom(C, A)$$ where the end turned into the coend and the exponent turned into the product. In our case, we may reduce by Yoneda $\hom(C, A\otimes B)$ with $\hom(-, C \otimes D)$ under the coend: $$\int^C \hom(-, C \otimes D) \times \hom(C, A\otimes B) \approx \hom(-, (A \otimes B) \otimes D)$$ obtaining: $$\int^{A, B, D} F(A) \times G(B) \times H(D) \times \hom(-, (A \otimes B) \otimes D)$$ Due to associativity of $\otimes$ we know that $(A \otimes B) \otimes D \approx A \otimes (B \otimes D)$. Now it suffices to "unwind" the coend in the other direction.


The category of cocontinuous functors on $\widehat{C}$ with values in a cocomplete category is equivalent to the category of functors on $C$. In order to construct a natural isomorphism $(F \otimes G) \otimes H \cong F \otimes (G \otimes H)$ for $F,G,H \in \widehat{C}$, it is therefore enough to treat the case $F \in C$. Similarily, one reduces to $G \in C$ and $H \in C$. But then we use the associativity in $C$.

  • $\begingroup$ I am a bit confused by your answer (it may be the case that I do not fully understand you). Clearly, what you have written is true. However, I think you need one additional fact to make the argument rigorous, and I am not sure if writing all these facts together is really different from the coend computation. $\endgroup$ – Michal R. Przybylek Feb 1 '13 at 23:47
  • $\begingroup$ No, it's completely elementary. $\endgroup$ – Martin Brandenburg Apr 14 '13 at 21:05

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